Sequences and series *: Meetkundige rijen *
Geometric series
A geometric sequence has the form #a_k=dc^{k-1}# with #k\in\mathbb{N}# and #c#, #d\in\mathbb{R}#.
The sequence hence starts at #k=1# when #d,\,dc,\,dc^2,\,dc^3,\ldots#
A geometric series originates by taking the sum #a_k=dc^{k-1}# on all the elements of a geometric sequence:\[b_n=\sum_{k=1}^na_k=\sum_{k=1}^ndc^{k-1}=d\sum_{k=1}^nc^{k-1}\tiny.\]
For #c\ne1# holds \[\sum_{k=1}^nc^{k-1}=\sum_{k=0}^{n-1}c^k=\frac{c^n-1}{c-1}\tiny.\]
We prove this formula with full induction. We need to prove the formula:\[\sum_{k=0}^{n-1}c^k=\dfrac{c^n-1}{c-1}\tiny.\]If #n=1#, then the left hand side is #c^0 = 1# and the right hand side #\dfrac{c^1-1}{c-1} = 1#. Because boths sides are equal, the conclusion is correct for #n=1#.
Now let #n\ge1# and assume that the formula is correct for #n#. For #n+1# then holds
\begin{eqnarray*}\sum_{k=0}^{n}c^k&=&\left(\sum_{k=0}^{n-1}c^k\right) + c^n= \left(\dfrac{c^n-1}{c-1}\right)+ c^n\\&=& \dfrac{c^n-1+c^n(c-1)}{c-1}= \dfrac{c^n-1+c^{n+1}-c^n}{c-1}\\&=& \dfrac{c^{n+1}-1}{c-1}\tiny.\end{eqnarray*}
This way we derived the formula for #n+1#. Because full induction the formula is correct for #n\ge1#.
If #c\neq1#, then #\sum_{k=0}^{n-1}c^k=\frac{c^n-1}{c-1}#. Entering of #n=6# and #c=3# gives\[\sum_{k=0}^{5}3^k=\frac{3^{6}-1}{3-1}=364\tiny.\]
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